Question

A vessel with water is placed on a weighing pan and it reads 600 g. Now a ball of mass 40 g and density 0.80 g cm⁻³ is sunk into the water with a pin of negligible volume, as shown in figure keeping it sunk. The weighting pan will show a reading(a) 600 g(b) 550 g(c) 650 g(d) 632 g

Answer 1

Answer:

(b) 550 g is the correct answer.

Explanation:

Given data:

A vessel with water is placed on a weighing pan.

Weight pf vessel = 600 g.

Now a ball of is sunk into the water with a pin of negligible volume.

mass of ball = 40 g

density 0.80 g cm^-3

It is shown in figure keeping it sunk.

To find:

The weighting pan will show a reading=?

Here, buoyant force will act.

Buoyant force is basically the upward force exerted by a liquid that ceases the weight of object in the liquid.

It is also called as upthrust.

It is necessary for the object to afloat in the liquid.

Here, the buoyant force will increase.

Thus weight will also increase.

Buoyant force is given by the formula;

F = Vρg

F = 40/0.80 * 1 *10

Let us assume g = 10m/s^2

F = 400/0.80

F = 50gm wt.

Reading of balance will increase.

= 500+50

= 550 gm wt.

Hence, b) 550 g is the correct answer.